[tex]\int _1^0\frac{x^2}{x+1} dx =\int \frac{\left(u-1\right)^2}{u}du =\int \:udu-\int \:2du+\int \frac{1}{u}du =\frac{u^2}{2}-2u+\ln \left|u\right|
\\
\\
\\ =\frac{\left(x+1\right)^2}{2}-2\left(x+1\right)+\ln \left|x+1\right|+C
\\
\\ =-\frac{3}{2}-\left(\ln \left(2\right)-2\right) =-\ln \left(2\right)+\frac{1}{2}[/tex]
